Question

A mixture of three gases (Ar, Ne, and CO2) has a total pressure of 1.2 atm. If the mixture of gases is composed of 25.0 g of each gas, what is the partial pressure of Ne?

Answer #1

Molar mass of Ar = 39.95 g/mol

Molar mass of Ne = 20.18 g/mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

n(Ar) = mass/molar mass

= 25.0/39.95

= 0.6258

n(Ne) = mass/molar mass

= 25.0/20.18

= 1.2389

n(CO2) = mass/molar mass

= 25.0/44.01

= 0.5681

n(Ar),n1 = 0.6258 mol

n(Ne),n2 = 1.2389 mol

n(CO2),n3 = 0.5681 mol

Total number of mol = n1+n2+n3

= 0.6258 + 1.2389 + 0.5681

= 2.4327 mol

p(Ne),p2 = (n2*Ptotal)/total mol

= (1.2389 * 1.2)/2.4327

= 0.61 atm

Answer: 0.61 atm

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